Question 1
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Vector quantities have both magnitude and direction, while scalars have only magnitude.
In option A, distance and speed are scalars, so it is incorrect.
Option C lists energy, mass, and temperature, all of which are scalar quantities.
In option D, time is a scalar, so it does not consist solely of vectors.
Electric field strength, momentum, and weight all possess direction and magnitude, making option B the correct choice.
Therefore, only option B contains exclusively vector quantities.
Question 2

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
To return to the third floor, the total displacement must be zero (area above time axis = area below time axis).
From P to Q, the area is positive (upward displacement); from Q to R, velocity is zero (stopped, still above third floor).
From R to S, the area is negative, cancelling some upward displacement; by S, the lift has returned to the starting point (third floor).
Between S and T, the velocity is zero again, meaning the lift is stationary at the third floor during this interval.
Therefore, the elevator is on the third floor again during the interval S to T only.
Question 3
Which row correctly compares the object on the planet to the object on the Earth?

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Mass is the quantity of matter in an object and does not change with location, so mass on planet = mass on Earth.
Weight is the gravitational force (W = mg). Since the planet’s gravitational field strength (g) is twice that of Earth’s, the weight is doubled.
Therefore, the correct comparison is: mass same, weight greater, which corresponds to row A.
Question 4

Each liquid is poured, in turn, into a container of water. The density of water is 1000 kg / m³.
Which liquids rise to the surface?
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
A liquid floats on water if its density is less than that of water (1000 kg/m³).
From the data, liquid X has a density of 940 kg/m³, which is less than water, so it rises to the surface.
Liquid Y (1250 kg/m³) and liquid Z (1800 kg/m³) have densities greater than water, so they sink.
Since only X has a density lower than water, option A is correct.
Question 5
The diagram shows a car moving along a road.
The force due to the engine is 1500 N and the total drag force is 200 N.

What is the motion of the car?
A. decreasing speed forward
B. increasing speed forward
C. reversing
D. constant speed
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
The resultant force is the difference between the forward engine force (1500 N) and the backward drag force (200 N).
This gives a resultant force of 1300 N acting in the forward direction of the car’s motion.
According to the syllabus, a resultant force acting on an object will cause it to accelerate (change its speed).
Since the resultant force is in the same direction as the car’s movement, the speed will increase.
Therefore, the car is increasing speed forward, making option B the correct choice.
Question 6


Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
To maximize cutting force, the rope should be placed as close to the pivot as possible (position Y) to reduce the distance and increase the force exerted.
The hands should apply effort at the farthest point from the pivot (point Q) to maximize the perpendicular distance.
This arrangement creates the largest possible moment for the same applied effort due to the principle of moments.
Therefore, rope at Y and hands at Q produces the greatest mechanical advantage and cutting force.
Question 7
Its velocity changes from 6 m / s to 10 m / s.
Which impulse has acted on the object?
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Impulse is defined as the change in momentum, given by the equation impulse = Δ(mv) = m(v – u).
Substituting the given values: mass m = 20 kg, initial velocity u = 6 m/s, and final velocity v = 10 m/s.
The change in velocity is (10 – 6) = 4 m/s, so the impulse = 20 kg × 4 m/s = 80 kg m/s, which is equivalent to 80 N s.
Therefore, the impulse acting on the object is 80 N s.
Question 8
Four examples of energy resources are listed.
- water waves
- geothermal
- wind
- water behind hydroelectric dams
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Water waves, wind, and water behind hydroelectric dams all ultimately derive their energy from the Sun’s radiation.
Solar heating drives the water cycle (evaporation, cloud formation, rain) and creates temperature differences that cause wind.
Geothermal energy, however, comes from heat generated within the Earth’s core and does not depend on the Sun.
Therefore, three of the four listed resources depend on radiation from the Sun, making option C the correct answer.
Question 9

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
The decrease in gravitational potential energy is given by the formula ΔEₚ = mgΔh.
From the diagram, the vertical height Δh between P and Q is 5.0 m.
Substituting the values: m = 2.0 kg, g = 9.8 N/kg, and Δh = 5.0 m, we get ΔEₚ = 2.0 × 9.8 × 5.0 = 98 J.
This calculation shows the energy transferred from the gravitational store to other forms as the box descends.
Options A, C, and D do not match the product of these specific values for mass, field strength, and height.
Therefore, option B is the correct answer.
Question 10


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▶️ Answer/Explanation
Detailed solution:
Gravitational potential energy (GPE) peaks at the top of the jump and is zero at take-off and landing.
Kinetic energy (KE) is maximum at P, decreases to a minimum at the peak as speed drops, then increases again towards R.
Due to air resistance, the total mechanical energy decreases, so KE at R is less than at P.
Graph B correctly shows GPE rising and falling symmetrically while KE dips and recovers to a lower value.
The other graphs either show incorrect energy variations or fail to account for energy loss due to air resistance.
Question 11
At the bottom of which tank is the pressure due to the water greatest?

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
The pressure exerted by a liquid at the bottom of a tank depends on the vertical depth of the liquid.
Comparing the four diagrams, the water level is highest in tank A, indicating the greatest depth.
Since the liquid is the same (water, so density is constant) and gravity is constant, pressure increases with depth.
The shape or width of the tank does not affect the pressure at the bottom; only the height of the water column matters.
Therefore, the pressure due to the water is greatest at the bottom of tank A.
Question 12
The volume of the container does not change.
What happens to the particles of the gas?
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Heating a gas increases the temperature, which corresponds directly to an increase in the average kinetic energy of the particles.
Since the container is sealed and rigid, the volume is constant, meaning the particles cannot move further apart on average.
The mass and individual volume of the gas particles remain unchanged during this physical process.
Therefore, option B correctly identifies the change in the motion of the particles.
Question 13

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Since the temperature remains constant, the average speed of the air particles does not change.
Compressing the syringe reduces the volume, meaning the particles have less space to move.
This results in the particles striking the walls of the container more frequently per unit time.
According to the kinetic particle model, this increase in collision frequency causes the pressure to rise.
Therefore, option C is correct because the pressure increases due to more frequent collisions, not faster particles.
Question 14
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Evaporation occurs when the most energetic particles at the surface of a liquid overcome attractive forces and escape into the air.
Since the highest kinetic energy particles leave, the average kinetic energy of the remaining particles in the liquid decreases.
A decrease in average kinetic energy corresponds directly to a drop in the temperature of the liquid.
This explains why sweating cools the body; the escaping high-energy water molecules leave behind cooler liquid.
Therefore, the correct description is that the most energetic particles escape, causing the temperature to decrease.
Hence, option C is the correct statement regarding evaporation.
Question 15
Which effect does this have on its mass and on its density?

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
When a solid is heated, its mass remains unchanged because the number of particles does not change.
Thermal expansion causes the solid’s volume to increase while its mass stays constant.
Since density is defined as mass per unit volume (\(\rho = m/V\)), an increase in volume with constant mass results in a decrease in density.
Therefore, the mass stays the same and the density decreases, which corresponds to option C.
Question 16

Which row explains why the water at the bottom of the test-tube remains cold?

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Heating water at the top makes it less dense, so it stays at the top and does not sink.
Convection currents cannot form because the heated, less dense fluid is already above the cooler, denser fluid.
Water is a poor conductor of heat, so thermal energy does not travel downwards effectively to warm the bottom.
Since there is no circulation and minimal conduction, the bottom water remains cold despite boiling at the surface.
Therefore, row D correctly states that water is a poor conductor and convection does not occur.
Question 17
In two seconds, three wave crests spread out on the surface of the water.

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
The concentric circles shown in the diagram represent lines joining points of the wave that are in phase, known as wavefronts.
Distance X measures from crest to trough, which is twice the amplitude, so A is incorrect.
Distance Y measures between two wavefronts, which is indeed the wavelength, making B a plausible distraction, but C is the universally correct description of the circles.
Frequency is the number of waves per second; 3 crests in 2 seconds gives 1.5 Hz, not 3 Hz, so D is incorrect.
Therefore, the statement that each circle represents a wavefront is the definitively correct answer.
Question 18
When they arrive, they cause the ground to vibrate with a frequency of 2.0 Hz.
What is the wavelength of these seismic waves?
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
First, calculate wave speed: v = distance / time = 300 km / 100 s = 3.0 km/s.
Use the wave equation v = fλ, rearranged to λ = v / f = 3.0 km/s ÷ 2.0 Hz.
This yields a wavelength of 1.5 km, matching option B.
Question 19

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Diffraction is the spreading of waves around an obstacle or through a narrow gap, which is clearly shown in diagram A.
In diagram A, the straight wavefronts bend and spread out into a semicircular pattern after passing through the opening.
Diagram B shows reflection, where waves bounce off a surface, while diagram C illustrates refraction due to a change in depth.
Diagram D also depicts refraction as the waves change direction without the characteristic spreading seen in diffraction.
Therefore, only diagram A correctly illustrates the phenomenon of diffraction of water waves.
Question 20
The diagram shows the ray of light incident on mirror 1.

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
The ray strikes mirror 1 at 50° to the surface, so the angle of incidence is 90° – 50° = 40°.
By the law of reflection, the angle of reflection at mirror 1 is also 40°.
Using geometry of the right-angled triangle formed by the mirrors, the ray strikes mirror 2 at an angle of incidence of 70°.
Applying the law of reflection again, the angle of reflection at mirror 2 is also 70°.
Therefore, option C is the correct answer.
Question 21

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
A converging (convex) lens brings parallel rays of light together to a single point called the principal focus.
Diagram A shows diverging rays, which is the effect of a concave (diverging) lens on a parallel beam.
Diagram C shows rays passing straight through without bending, which occurs only when passing through the optical center.
Diagram D shows parallel rays remaining parallel after the lens, which is not physically accurate for either type of lens.
Therefore, only diagram B correctly illustrates the converging action of a convex lens on incident parallel light rays.
Question 22
- sterilising water
- optical fibres
- satellite television signals
- mobile (cell) phones
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Ultraviolet (UV) radiation is used for sterilising water due to its germicidal properties, effectively killing bacteria and viruses.
Infrared (IR) waves are commonly used in optical fibres for telecommunications because they can carry high rates of data with minimal loss.
Microwaves are used for both satellite television signals and mobile (cell) phone communications as they can penetrate the atmosphere and walls.
Radio waves and visible light are not involved in all three specific applications listed in the question.
Therefore, the correct combination matching the applications is ultraviolet, infrared, and microwaves.
Question 23
How long after making the sound does the man hear the echo?
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
The sound travels from the man to the wall and back, covering a total distance of 2 × 110 m = 220 m.
Using the equation time = distance / speed, the time taken is 220 m / 330 m/s = 0.666… s.
This rounds to 0.67 s, which matches option B.
The calculation requires doubling the one-way distance because an echo involves reflection back to the source.
Options A, C, and D result from either using only the one-way distance or incorrect arithmetic.
Therefore, the echo is heard 0.67 s after the sound is made.
Question 24

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
The strength of a magnetic field is indicated by the spacing of the field lines; closer lines mean a stronger field.
At point P (near the pole), the lines are close together, indicating a strong field.
As the path moves towards R (the midpoint between poles), the lines spread out, showing the field weakens to a minimum.
Moving from R to Q (approaching the other pole), the lines converge again, so the field strength increases.
Therefore, the field strength is strong at P, decreases to a minimum at R, and increases towards Q.
Question 25

Which diagram shows the direction of the force acting on the electron due to the charged sphere?

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
The electric field lines point away from the sphere, indicating it has a positive charge.
By convention, the direction of an electric field is the force on a positive test charge.
An electron has a negative charge, so it experiences a force in the opposite direction to the field.
At point Y, the field is directed to the right, therefore the force on the electron is towards the left.
This matches the arrow direction shown in option D, pointing back toward the positive sphere.
Question 26
What is the electromotive force (e.m.f.) of the cell?
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Electromotive force (e.m.f.) is defined as the energy transferred per unit charge.
The formula is E = W / Q, where W is the energy transferred in joules and Q is the charge in coulombs.
Substituting the given values, E = 12 J / 3.0 C = 4.0 J/C.
Since 1 J/C is equal to 1 volt (V), the e.m.f. of the cell is 4.0 V.
Therefore, the correct answer is option B.
Question 27
The putty is now moulded into a cylinder of diameter 2d that has the same volume.

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Resistance R = ρL/A. Doubling diameter increases cross-sectional area (A ∝ d²) by factor of 4.
Constant volume means Area × Length is constant, so length L decreases by factor of 4.
New resistance R’ = ρ(L/4) / (4A) = (1/16) × (ρL/A) = R/16.
The resistance therefore decreases by a factor of 16, making option D correct.
Question 28
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
A diode is a semiconductor device that acts as a one-way valve for electric current.
It has very low resistance in forward bias, allowing current to flow, and very high resistance in reverse bias, blocking current.
Option A describes a thermistor, option B describes a fuse, and option D describes a relay.
Therefore, the defining characteristic of a diode is that it only permits current to flow in a single direction.
Hence, option C is the correct answer.
Question 29
Which arrangement of resistors will give an overall resistance of 25 Ω?

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
For two 10 Ω resistors in parallel, the combined resistance is (10×10)/(10+10) = 5 Ω. Adding this parallel pair in series with two separate 10 Ω resistors gives 10 + 10 + 5 = 25 Ω.
Option A gives 10 Ω in parallel with 10 Ω (5 Ω) and series with 10 Ω = 15 Ω. Option B gives three 10 Ω in parallel = 3.33 Ω. Option C gives two parallel pairs (5 Ω each) in series = 10 Ω.
Only arrangement D provides the correct combination of series and parallel connections to achieve exactly 25 Ω total resistance.
Question 30
The circuit diagram shows a thermistor in a potential divider.
Two voltmeters X and Y are connected in the circuit.

When the temperature of the thermistor increases, its resistance decreases.
As the thermistor becomes warmer, what happens to the reading on each voltmeter?

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
As temperature increases, the thermistor’s resistance decreases, so its share of the total potential difference drops.
Voltmeter X is across the fixed resistor, which now has a larger proportion of the supply voltage, so its reading increases.
Voltmeter Y is across the thermistor, and since its resistance and p.d. have decreased, the reading on Y decreases.
This behavior follows the potential divider principle, where the voltage is shared in proportion to the resistances.
Hence, the reading on voltmeter X increases, and the reading on voltmeter Y decreases.
Question 31
In an experiment, strong magnetic fields provide a force that keeps charged particles moving with a horizontal circular motion.
Positive particles are accelerated clockwise when viewed from above as shown.

In which direction is the magnetic field?
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Using Fleming’s left-hand rule, the force (centripetal, towards the centre) is given by the thumb, the first finger is the magnetic field, and the second finger is the current (direction of positive particle motion).
For clockwise motion viewed from above, the velocity at the top is directed to the right; pointing the second finger right and thumb towards the centre requires the first finger to point vertically upwards, out of the page.
Thus, the magnetic field must be directed vertically out of the page to produce the observed clockwise circular path for positive particles.
Question 32
How does the magnetic field change if the current is increased?

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
The strength of the magnetic field around a solenoid is directly proportional to the current flowing through it.
Increasing the current increases the magnitude of the magnetic field, which is represented by a greater density of magnetic field lines.
The pattern and direction of the field remain the same, but the field becomes stronger everywhere around the solenoid.
Option A correctly shows the same field pattern with more lines, indicating a stronger field.
Options B and D incorrectly change the direction of the field, while Option C shows no change in field strength.
Therefore, Option A is the correct choice.
Question 33


Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
For maximum efficiency, the core should be laminated soft iron to reduce eddy current losses and provide easy magnetization.
The wire should be thick copper to minimize resistive heating from the current flowing through the coils.
Row A correctly identifies laminated soft iron for the core material and thick copper for the wire material.
Other rows propose materials (e.g., steel or thin wire) that would increase energy losses through heat or magnetic hysteresis.
Therefore, option A describes the best construction conditions for a highly efficient transformer.
Question 34
What is the nature of the products?
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Radioactive decay involves an unstable nucleus emitting particles or energy to become more stable.
During α-decay or β-decay, the proton number changes, resulting in a nucleus of a different element.
Gamma radiation is often emitted alongside alpha or beta particles to release excess energy.
Therefore, the correct description is a new element’s nucleus formed with both α-particle and γ-radiation emitted.
This matches the behavior of isotopes undergoing alpha decay with accompanying gamma emission.
Question 35
What causes the proton number to decrease by 1?
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
In α-decay, the proton number decreases by 2 and the nucleon number decreases by 4.
In β-decay, a neutron changes to a proton, so the proton number increases by 1.
γ-decay does not change the proton or nucleon number.
Therefore, an α-decay (proton number decreases by 2) followed by a β-decay (proton number increases by 1) results in a net decrease of 1 in the proton number.
Option A correctly describes this sequence of decays to achieve the required change.
Question 36
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Half-life is defined as the time taken for half the unstable nuclei in a sample to decay, which reduces the count rate and mass proportionally.
Options A, B, and C are all valid descriptions of half-life in terms of count rate, nuclei number, or mass of the parent isotope.
Option D is incorrect because half-life refers to the decay of unstable nuclei, not specifically the decay of neutrons alone.
Neutron decay is just one possible process (beta decay), but half-life applies to the overall nuclear transformation of the isotope.
Therefore, the statement in D is not a correct definition of half-life.
Question 37
The graph shows how the count rate at the detector varies with the thickness of the paper.
The rollers squeeze the paper when the count rate at the detector drops below 12 counts per second and then stop squeezing when the count rate rises above 13 counts per second.

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
The control system maintains the count rate between 12 and 13 counts per second.
From the graph, a count rate of 12–13 counts/s corresponds to a paper thickness range of approximately 0.50 mm to 0.60 mm.
Since the rollers adjust the thickness to keep the count rate within this window, the output paper thickness will lie in this range.
Therefore, the thickness of the paper coming out of the machine is between 0.50 mm and 0.60 mm.
Question 38

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Mars is the fourth planet from the Sun, located between Earth (365 days) and Jupiter (4300 days).
According to the table, the orbital duration increases with distance from the Sun.
Since Mars is further from the Sun than Earth but closer than Jupiter, its orbital period must be greater than 365 days but significantly less than 4300 days.
690 days (approximately 1.9 Earth years) fits logically within this range based on Kepler’s third law.
Therefore, option B is the correct orbital duration for Mars.
Question 39
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
The Sun produces energy through the process of nuclear fusion, specifically the fusion of hydrogen nuclei into helium.
Fission involves splitting heavy nuclei, which is not the process occurring in the Sun’s core.
Uranium is not involved in solar reactions; the Sun primarily consists of hydrogen and helium.
In stable stars like the Sun, the inward gravitational force is balanced by the outward pressure from fusion energy.
Thus, the correct answer is the fusion of hydrogen, as described in the nuclear reactions powering the Sun.
Question 40
What is this distance in metres?
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
1 light-year is defined as the distance light travels in one year, which is \(9.5 \times 10^{15}\) m.
The diameter of the Milky Way is given as 100,000 light-years.
To convert this to metres, multiply the distance of one light-year by 100,000: \(100,000 \times 9.5 \times 10^{15} \mathrm{m}\).
This calculation yields \(9.5 \times 10^{5} \times 10^{15} = 9.5 \times 10^{20} \mathrm{m}\).
Therefore, option D correctly represents the diameter of the Milky Way in metres.
